# Any ideas for the following limit of partial sums of binomial coefficients?

Let $$a$$ be an odd integer $$≥3$$. It appears that: $$\lim_{n\rightarrow\infty}\frac{1}{2^{n}}\sum_{m=0}^{\left\lfloor n\frac{\ln2}{\ln a}\right\rfloor }\binom{n}{m}=\begin{cases} 1 & \textrm{if }a=3\\ 0 & \textrm{if }a\geq5 \end{cases}$$

Any ideas as to how to prove this? I tried using Stirling's formula, but everything became horrible.

• law of large numbers: the sum of binomials up to $\alpha\cdot n$ is $(1-o(1))2^n$ if $\alpha>1/2$ and $o(2^n)$ if $\alpha<1/2$. – Fedor Petrov Aug 28 '19 at 22:53

Fix $$0<\alpha<\frac12$$ and consider the sum $$\sum_{0\le m\le\alpha n} \binom nm$$. (For simplicity let's assume $$\alpha n$$ is an integer.) The ratio of the $$m$$th term to the $$(m+1)$$st term in this sum is at most $$\alpha/(1-\alpha)$$; this means that the sum is bounded by $$\binom n{\alpha n} \sum_{k=0}^\infty \bigg(\frac\alpha{1-\alpha}\bigg)^k = \binom n{\alpha n} \frac{\alpha-1}{2\alpha-1},$$ which is $$o(2^n)$$ by Stirling's formula or probably a simpler argument (an upper bound for the central binomial coefficient will suffice). A similar argument works for the upper tail $$\sum_{\beta n\le m\le n} \binom nm$$ when $$\frac12<\beta<1$$.